Question

A gramophone disc is rotating at 78 rotations per minute.Due to power cut, it comes to rest after 30 s .The angular retardation of the disc will be:
A 0.27 $radians/ sec^2$
B 0.127 $radians/ sec^2$
C 12.7 $radians / sec^2$
D 0 $radians / sec^2$

Answer 1

Use the concept of angular acceleration and angular velocity.The angular acceleration is also known as rotational acceleration. It is a quantitative expression of the change in angular velocity per unit time. The acceleration vector, magnitude, or length is directly proportional to the rate of change in angular velocity.

Complete step by step answer:
Angular acceleration ($\alpha$ )=(${\omega_f}$ -${\omega_i}$)/t
Where, ${\omega_f}$ is the final angular velocity of the gramophonic disc, ${\omega_i}$ is the initial angular velocity of the gramophonic disc and $t$ is the time.
$\omega$=2$\dfrac{\pi}{t}$ or 2 $\pi f$
Where $f$ is the frequency.
It is given to us that the disc makes 78 rotations in 1 minute.
So the frequency (f)=$\dfrac{78}{60}$ [frequency is the number of rotations made by the disc in 1 second]
Time is 30 seconds. So, angular acceleration ($\alpha$) =0-2 $\pi$ f/t
$\alpha =- 2 \times\pi \times\dfrac{78}{60} \times 30$
$\therefore \alpha = -0.272 radians/s^2$

Hence, option (A) is the correct answer.

Note: The negative symbol in the angular acceleration represents that the disc is retarding which means acceleration is in opposite to the direction of motion of the disc.The vector direction of the acceleration is perpendicular to the plane where the rotation takes place. Increase in angular velocity clockwise, then the angular acceleration velocity points away from the observer. If the increase in angular velocity counterclockwise, then the vector of angular acceleration points toward the viewer.